GATE 2020 EC Q.20: solution and answer

GATE 2020 · ECQ.20 · 1 mark · NAT

A 10-bit D/A converter is calibrated over the full range from 0 to 10 V. If the input to the D/A converter is 13A (in hex), the output (rounded off to three decimal places) is ________ V.

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Answer and solution

ANSWER

The D/A output is about 3.069 V ( $314 \times \frac{10}{1023}$ V).

Official answer: 3.050 to 3.080Source: Official answer key, GATE 2020 EC (IIT Delhi)

Convert the hex input to decimal: $(13 A)_{16} = 1 \times 1 6^{2} + 3 \times 16 + 10 = 256 + 48 + 10 = (314)_{10}$ .
Calibrated over the full range 0 to 10 V means the all-ones code $(1111111111)_{2} = 2^{10} - 1 = 1023$ gives the full-scale output of $10 V$ , and code 0 gives $0 V$ .
Resolution (step size) $= \frac{V _{F S}}{2 ^{n} - 1} = \frac{10}{1023} V \approx 9.775 mV$ .
Output of a DAC $= (decimal input) \times (step size)$ : $V_{o} = 314 \times \frac{10}{1023} = \frac{3140}{1023} \approx 3.0694 V$
Rounded to three decimal places: $V_{o} \approx 3.069 V$ .

KEY IDEA

A DAC's output is the decimal value of the input code times one step. "Calibrated over the full range 0 to 10 V" means the largest code, $2^{n} - 1$ , gives 10 V, so one step is $\frac{10}{2 ^{n} - 1}$ V. Convert the hex input to decimal first, then multiply by the step.

WATCH OUT FOR

Many students read 13A as decimal 13 or 130, or mis-convert A (it is 10, so 13A in hex is 314 in decimal). Some divide by $2^{10} = 1024$ instead of $2^{10} - 1 = 1023$ . That gives $314 \times \frac{10}{1024} \approx 3.066$ V, which also falls inside the official range of 3.050 to 3.080, so the key accepts both conventions. Under the stated calibration (all-ones code = 10 V), 1023 is the correct divisor.

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GATE 2020 EC Q.20: solution and answer

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